Question

The work function of metals A and B are in the ratio 1:2. If light of frequencies $f$ and $2f$ are incident on metal surfaces A and B respectively, the ratio of the maximum kinetic energies of the photo electrons emitted is :

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is B $1:2$

The maximum kinetic energy of photoelectrons is

$E_{max}=h\nu -\varphi$

where, $\nu$ is the frequency of incident light and $\varphi$ is photoelectric work function of metal.

As per the problem, for metal A,

$E_{Amax}=h\nu -{\varphi }_A$

and

for metal B,

$E_{Bmax}=h\left(2\nu \right)-{\varphi }_B$

Now,

$\frac{E_{Amax}}{E_{Bmax}}=\frac{h\nu -{\varphi }_A}{2h\nu -{\varphi }_B}$

$\frac{E_{Amax}}{E_{Bmax}}=\frac{\frac{h\nu }{{\varphi }_A}-1}{\frac{2h\nu }{{\varphi }_A}-\frac{{\varphi }_B}{{\varphi }_A}}$

$\frac{E_{Amax}}{E_{Bmax}}=\frac{\frac{h\nu }{{\varphi }_A}-1}{\frac{2h\nu }{{\varphi }_A}-2}$ ..................(since $\frac{{\varphi }_A}{{\varphi }_B}=\frac{1}{2}$)

$\frac{E_{Amax}}{E_{Bmax}}=\frac{\frac{h\nu -{\varphi }_A}{{\varphi }_A}}{\frac{2h\nu -2{\varphi }_A}{{\varphi }_A}}$

$\frac{E_{Amax}}{E_{Bmax}}=\frac{h\nu -{\varphi }_A}{2(h\nu -{\varphi }_A)}$

$\frac{E_{Amax}}{E_{Bmax}}=\frac{1}{2}$